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精品资源欢迎下载“分式的乘除法复习题填空题1.约分:2m 2m 1a3 13n2a2.1.2.3.1 -m计算:_4(2c)(-x)2n 1（n为正整数）ab2c)2b23c24ab3(-xy2)2x112x11bc2a2；(xy-x2)-2xyxy判断题下列运算正确的打，错误的打“x2x y x y 2x 2x=1 =x y x y y x y x y(J3(d)26 3x y3z9 4x y2zb2b 4n4.()2n=-F (n为正整数)( aa5.2b3 38b93a2)-27a6三.选择题1.已知x:=3则分式(x-y)2的值是（2.A. 34在分式B.2726C.D.1413a3x6xy2x 2y2 x2 x- yy2y _ x(x - y)2中，最简分式有（个 D. 4 个A. 1 个 B. 2 个 C. 33.下列各式正确的是（）A.22x - yx - y-22 二x yx yB.2x 2x 1 x 1 22= 一(/2x - x -1x -1,2a a b, c 、2 cD.() = -2b 2ba b a b四.计算1.c 2 / 3x 、 ， 6z 、8x (-3) (- -)4y x y2.2/ 3x 33,3.()(a -b )(a bb -a 2)2b ay2 (a b)y ab a2 x2 3. x2 - (a b)x ab b2 - y24.x2。6x 9 一 x2。9x 3一Tx2 -x-6x2 -3x -10 2(x -5)5.2x -6x2x - 64T4T? (x 3) T6.2.3x x 11 - x-22x 11 x一(x2 x -2)7.a2 7a 10 a3 1 a2 - a 1+-2_2_a 6a 5 a 4a 4 a 28.22a -bab)29.(a2 一 x2)3 (a2 2ax x2)21a2 x2a4 - x4(a2 - ax x2)2a2b2-c2-2ab , a2-b2-c22bca2b2 c2-2ac a2 b2c22ab11.(x y)22 xy -y212.2 3y x 2-3yx1 yx -3211x - y3213.32x3 2x2 x 2x2 4x 422,x -4x 4. x 1x3 -2x2 - x 2 x2 714.2x -16.12 -4x72x -4x 4 x -415.a7b2 i (a2 -b2)4 . ra2(b-a)n3 一37 16._ 2_ 2(8a -2b )-4a2b 4ab2 b32_22b - 5ab - 3ab2，其中 a =1，b = -ob 一 5ab 6a24【试题答案】1.+1;2.2 4 24a2b4c2-1;27a9T522x y- yy-x.1.2.3.4.X 5.1. C2. B3. B四.1.解:原式= 8x23x4y36z-2-x y36xzy2.解:原式27x9(a b)3222 (a b)(a 一b)(a ab b )匕(b -a2)222927(a -b)(a ab b )x(a b)(b - a2)23.解:原式(y a)(y b) (a x)(a -x)(x -a)(x -b) (b y)(b - y)(y a)(x a)(x -b)(b - y)(y a)(x a) (x -b)(y -b)4.解:原式2(x-3)2(x -3)(x 2)(x-5)(x 2)(x 3)(x-3)x 3 _ 12(x -5)25.解:原式2(x -3) 12 Z(x -2) x 3(x 3)(x-2)6.解:原式7.解:原式8.解:原式9.解:原式3-xx-2x2 x 1(1 - x)(x 2 x 1)1 x22_x 1 (1 x )(x 2)x2 1x2 x 1(x 2)(x-1)=-(x 2) = -x -2(a 2)(a 5)(a 1)(a 5)2(a 1)(a - a 1)(a b)2(a -b)2(a x)3(a -x)3(a 2)23a(a-b)3a3(a b)2a(a b)2b2(a2 x2)(a x)(a x)(a x)410.解：原式(a b c)(a b c)(a b c)(a b c)(a b c)(a b c) a b c (a b c)(ab c) a bc11.解：原式22(x y) (x - y) (x y)(x -y)I T /7 2y(x - y) y(x y)y2x 3y 2x-3y12.解：原式=x ,6=2x+3y 2x-3y =12x -3y 2x 3y 2x -3y 2x 3yx 613.2_ 2解.原式=x(x 2) (x 2) (x-2)(x 1)(x-1) (x 2)2x2(x -2) - (x-2)x2 1(x2 1)(x 2) (x - 2)2 (x 1)(x-1)22(x 2)2 (x 1)(x -1)(x -2) x2 1x-2一 x 214.解：原式2(x -8) (x 2)(x -2)(x-2)24(3-x)1_ x -8x 2 - 2(3-x)(x-2)15.解：原式a7b2(a b)4 (a - b)488(a b)3b2(a - b)3(a b)a2a6(b - a)33a216.2(2a1b2(a 2b)2a b2a 4b2a b112x(-1) 4 -24=0112 ()-2411当a = - , b =一时，原式=24代数式的值为0
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